An electromagnetic wave of wavelength ‘λ’ is incident on a photosensitive surface of negligible work function. If ‘m’ mass is of photoelectron emitted from the surface has de-Broglie wavelength λ_d, then :

To find the relationship between the wavelength of the incident electromagnetic wave ($\lambda$) and the de-Broglie wavelength of the emitted photoelectron (${\lambda }_d$), we can follow these steps:

Step 1: Determine the energy of the incident photon
The energy ($E$) of an incident photon of wavelength $\lambda$ is given by Planck's relation:

$E=\frac{hc}{\lambda }$

where $h$ is Planck's constant and $c$ is the speed of light.

Step 2: Relate photon energy to photoelectron kinetic energy
According to Einstein's photoelectric equation, the maximum kinetic energy ($K$) of the emitted photoelectron is:

$K=E-\Phi$

where $\Phi$ is the work function of the photosensitive surface. Since the work function is negligible ($\Phi \approx 0$), the kinetic energy simplifies to:

$K=\frac{hc}{\lambda }$

Step 3: Relate kinetic energy to de-Broglie wavelength
The de-Broglie wavelength (${\lambda }_d$) of a photoelectron of mass $m$ and kinetic energy $K$ is given by:

${\lambda }_d=\frac{h}{\sqrt{2mK}}$

Step 4: Express the relation in terms of λ
Squaring both sides of the de-Broglie wavelength equation:

${\lambda }_d^2=\frac{h^2}{2mK}$

Now, substitute the value of $K=\frac{hc}{\lambda }$ into the equation:

${\lambda }_d^2=\frac{h^2}{2m\left(\frac{hc}{\lambda }\right)}$

${\lambda }_d^2=\frac{h\lambda }{2mc}$

Step 5: Rearrange to solve for λ
By rearranging the formula to express $\lambda$, we obtain:

$\lambda =\left(\frac{2mc}{h}\right){\lambda }_d^2$